Calculation of fault level at secondary side of transformer 5

[AusLee]

Hi,

I have little experience at HV side. Usually for my 800 kVA 11kV/400V 4% Ucc transformer I would calculate the fault at the LV side assuming an infinite bus at the primary, giving Isc = 800/3/230/0.04 = 29 kA

However in this particular instance we need to calculate the actual fault not this upper limit, and all we have as info is 3 phase fault level at primary is 7 kA.

How can I then calculate the fault level at the secondary, any additional info needed?

Thanks.

 

[sibeen]

Auslee, generally the upstream impedance doesn't make all that much difference, although I do tend to add it to my calculations and make it a *hand waving* 500 MVA. In this case you're given some figures, and the fault current is actually quite low.

So, work out the upstream capability, Short circuit capability:

Ssc = (11000/ root 3)* 7000 * 3 = 133 MVA -> [phase to line voltage * current * 3 phases]

You now need to find the upstream impedance

HV = 11000 Ssc = 133 MVA

Zup = (HV^2)/Ssc = 0.907 -> the upstream impedance in ohms

Now need to convert upstream impedance to the 400 volt base:

Zup_LV = Zup * (400/11000)^2 = 0.0012

That gives an answer in ohms, so to continue we need to convert the % impedance of the transformer into an ohmic value.

Ztr = (4/100)*(400^2/800000) = 0.008 -> (percentage/100) * (Line_line voltage squared/transformer size)

So now to work out the short circuit current:

Isc = 400/[(root 3 * (0.0012 + 0.008)] = 25.1 kA -> Voltage L-L / [root 3 * (Zup_LV + Ztr)].

And I hope that all makes sense

 

[rcw retired EE]

Try using the MVA method:
First: calculate the short circuit MVA of each piece of the system. (MVA at LV terminals if the equipment is connected to an infinite source)

Second- combine the MVA's. (MVA's combine like admittances, add when in parallel, useinvers in series).

Utility MVAsc= 11 kV x 7 kA x 1.732 = 133.4 MVA
Transformer MVA sc = MVA x 100%/%Z = 0.8 / 0.04 = 20 MVA

The two componenets are in series so we add them like admittances:

1/MVA system = 1/MVAutility + 1/MVAtx = 1/(1/133.4 + 1/20) = 1/ 17.4. MVA sc = 17.4 MVA

I sc = (17.4 MVA x 10^3) /(400V x 1.732) = 25.1 kA

 

[prc]

Let me suggest one more way.

Transformer Impedance = 0.04 pu

System impedance = Transformer MVA/ System fault MVA = 0.8/133.4 = 0.006 pu

Total impedance for fault = 0.04+0.006 =0.046

Fault MVA =0.8/0.046 =17.4 MVA

Isc= 17.4 X 10000/ 1.732x400 =25.1 kA

 

[cherryg]

One point to be noted. If you have still not ordered the Transformer, taking the minimum impedance using negative tolerance will give a onerous result.
Once the actual tested value of impedance is available, a more accurate short circuit value can be derived.

 

[ScottyUK]

In this case maximum impedance might be the more onerous result in terms of fault clearance time, given that the available fault level is quite low.

 

[wroggent]

Can you elaborate on your concern ScottyUK?

 

[mgtrp]

If the available fault current is low then protection may not be sensitive enough to operate, or may not operate as quickly, in the event of a fault. This could result in much more onerous duty being placed on equipment, e.g. as part of stability studies, arc flash analysis, etc

 

[wroggent]

I understood the basis of his comment but I just doubt it would be an issue. The OP didn't say anything about what system protection he has but let's just pretend it's a 1200A breaker - this should be reasonable for that transformer since 800kVA/(3^0.5*400V)=1156A. This thread was about bolted three phase faults which everyone agreed calculated out to 25.1kA. To my knowledge usually the instantaneous element of a breaker trips at around 10xFLA. 25.1kA is quite a bit higher than 12kA so I don't think he would have an issue with trip time in this case - the positive tolerance on top of the 4% he mentioned likely won't decrease the fault level that much. That's just what I was thinking when I read Scotty's comment.

 

[ScottyUK]

The intention was to say that the maximum fault level isn't always the worst case, and that the other extreme should be evaluated. I didn't check the numbers - lazy of me - but on a system with limited fault level it caught my eye as something to at least check.

Note to self : check the numbers before posting

 

[sibeen]

Scotty, I wouldn't beat myself up about it, it was a good point, well made. A lot of people just don't realise that a low fault level can be, in some cases, as bad as a high fault level.

To have that driven home can never be a bad thing, IMHO.

I must add, that is why I use the impedance method of calculation. As you progress down to the lower levels (boards) you just keep adding the impedances of the cabling, breakers, busburs etc, and get the resultant fault levels. This fault level can then be checked to see that distribution breakers aren't under-rated for the prospective fault current, but you can also check to make sure that they're able to trip in a timely manner.

As the above posts indicate, there is more then one way to skin a cat than the impedance method, but that's the one I learnt and is therefore the one I'm comfortable with.